electric charge properties of electric charge there are two kinds of electric charge. –like...

Electric Charge

Properties of Electric Charge

• There are two kinds of electric charge.– like charges repel– unlike charges attract

• Electric charge is conserved.– Positively charged particles are called protons.– Uncharged particles are called neutrons.– Negatively charged particles are called electrons.

Electric Charge

Properties of Electric Charge, continued

• Electric charge is quantized. That is, when an object is charged, its charge is always a multiple of a fundamental unit of charge.

• Charge is measured in coulombs (C).

• The fundamental unit of charge, e, is the magnitude of the charge of a single electron or proton.

e = 1.602 176 x 10–19 C

Electric Charge

Transfer of Electric Charge

• An electrical conductor is a material in which charges can move freely.

• An electrical insulator is a material in which charges cannot move freely.

Electric Charge

Transfer of Electric Charge, continued

• Insulators and conductors can be charged by contact.

• Conductors can be charged by induction.

• Induction is a process of charging a conductor by bringing it near another charged object and grounding the conductor.

Electric Charge

Transfer of Electric Charge, continued

• A surface charge can be induced on insulators by polarization.

• With polarization, the charges within individual molecules are realigned such that the molecule has a slight charge separation.

Coulomb’s Law

• Two charges near one another exert a force on one another called the electric force.

• Coulomb’s law states that the electric force is propor-tional to the magnitude of each charge and inversely proportional to the square of the distance between them.

Electric Force

1 22

2

charge 1 charge 2electric force = Coulomb constant

distance

electric C

q qF k

r

Coulomb’s Law, continued

• The resultant force on a charge is the vector sum of the individual forces on that charge.

• Adding forces this way is an example of the principle of superposition.

• When a body is in equilibrium, the net external force acting on that body is zero.

Electric Force

Sample Problem

The Superposition Principle

Consider three point charges at the corners of a triangle, as shown at right, where q1 = 6.00 10–9 C, q2 = –2.00 10–9 C, and q3 = 5.00 10–9 C. Find the magnitude and direction of the resultant force on q3.

Electric Force

Electric Force

Sample Problem, continuedThe Superposition Principle1. Define the problem, and identify the known

variables.Given:

q1 = +6.00 10–9 C r2,1 = 3.00 m

q2 = –2.00 10–9 C r3,2 = 4.00 m

q3 = +5.00 10–9 C r3,1 = 5.00 m

= 37.0º

Unknown: F3,tot = ? Diagram:

Electric Force

Sample Problem, continuedThe Superposition PrincipleTip: According to the superposition principle, the resultant

force on the charge q3 is the vector sum of the forces exerted by q1 and q2 on q3. First, find the force exerted on q3 by each, and then add these two forces together vectorially to get the resultant force on q3.

2. Determine the direction of the forces by analyzing the charges.

The force F3,1 is repulsive because q1 and q3 have the same sign.

The force F3,2 is attractive because q2 and q3 have opposite signs.

Electric Force

Sample Problem, continuedThe Superposition Principle3. Calculate the magnitudes of the forces with

Coulomb’s law.

–9 –9293 1

3,1 22 2

–83,1

–9 –9293 2

3,2 22 2

–93,1

5.00 10 C 6.00 10 CN m 8.99 10

( 3,1) C 5.00 m

1.08 10 N

5.00 10 C 2.00 10 CN m8.99 10

( 3,2) C 4.00m

5.62 10 N

C

C

q qF k

r

F

q qF k

r

F

Electric Force

Sample Problem, continuedThe Superposition Principle

4. Find the x and y components of each force.

At this point, the direction each component must be taken into account.

F3,1: Fx = (F3,1)(cos 37.0º) = (1.08 10–8 N)(cos 37.0º)

Fx = 8.63 10–9 N

Fy = (F3,1)(sin 37.0º) = (1.08 10–8 N)(sin 37.0º)

Fy = 6.50 10–9 N

F3,2: Fx = –F3,2 = –5.62 10–9 N

Fy = 0 N

Electric Force

Sample Problem, continued


5. Calculate the magnitude of the total force acting in both directions.

Fx,tot = 8.63 10–9 N – 5.62 10–9 N = 3.01 10–9 N

Fy,tot = 6.50 10–9 N + 0 N = 6.50 10–9 N

Electric Force



6. Use the Pythagorean theorem to find the magni-tude of the resultant force.

2 2 9 2 9 23, , ,

–93,

( ) ( ) (3.01 10 N) (6.50 10 N)

7.16 10 N

tot x tot y tot

tot

F F F

F

Electric Force



7. Use a suitable trigonometric function to find the direction of the resultant force.

In this case, you can use the inverse tangent function:

–9,

–9,

6.50 10 Ntan

3.01 10 N

65.2º

y tot

x tot

F

F

Coulomb’s Law, continued

• The Coulomb force is a field force.

• A field force is a force that is exerted by one object on another even though there is no physical contact between the two objects.

Electric Force

The Electric Field

Electric Field Strength

• An electric field is a region where an electric force on a test charge can be detected.

• The SI units of the electric field, E, are newtons per coulomb (N/C).

• The direction of the electric field vector, E, is in the direction of the electric force that would be exerted on a small positive test charge.

The Electric Field

Electric Field Strength, continued

• Electric field strength depends on charge and distance. An electric field exists in the region around a charged object.

• Electric Field Strength Due to a Point Charge

2

2

charge producing the fieldelectric field strength = Coulomb constant

distance

C

qE k

r

The Electric Field

Sample Problem

Electric Field Strength

A charge q1 = +7.00 µC is at the origin, and a charge q2 = –5.00 µC is on the x-axis 0.300 m from the origin, as shown at right. Find the electric field strength at point P,which is on the y-axis 0.400 m from the origin.


Electric Field Strength1. Define the problem, and identify the known

variables.Given:

q1 = +7.00 µC = 7.00 10–6 C r1 = 0.400 m

q2 = –5.00 µC = –5.00 10–6 C r2 = 0.500 m = 53.1º

Unknown:

E at P (y = 0.400 m)

Tip: Apply the principle of superposition. You must first calculate the electric field produced by each charge individually at point P and then add these fields together as vectors.

The Electric Field

Sample Problem, continuedElectric Field Strength2. Calculate the electric field strength produced by

each charge. Because we are finding the magnitude of the electric field, we can neglect the sign of each charge.

–69 2 2 51

1 2 21

–69 2 2 52

2 2 22

7.00 10 C8.99 10 N m /C 3.93 10 N/C

(0.400 m)

5.00 10 C8.99 10 N m /C 1.80 10 N/C

(0.500 m)

C

C

qE k

r

qE k

r

The Electric Field

Sample Problem, continuedElectric Field Strength3. Analyze the signs of the

charges.

The field vector E1 at P due to q1 is directed vertically upward, as shown in the figure, because q1 is positive. Likewise, the field vector E2 at P due to q2 is directed toward q2 because q2 is negative.

The Electric Field

Sample Problem, continuedElectric Field Strength4. Find the x and y components of each electric field

vector.

For E1: Ex,1 = 0 N/C

Ey,1 = 3.93 105 N/C

For E2: Ex,2= (1.80 105 N/C)(cos 53.1º) = 1.08 105 N/C

Ey,1= (1.80 105 N/C)(sin 53.1º)= –1.44 105 N/C

The Electric Field

Sample Problem, continuedElectric Field Strength5. Calculate the total electric field strength in both

directions.

Ex,tot = Ex,1 + Ex,2 = 0 N/C + 1.08 105 N/C = 1.08 105 N/C

Ey,tot = Ey,1 + Ey,2 = 3.93 105 N/C – 1.44 105 N/C = 2.49 105 N/C

The Electric Field

Sample Problem, continuedElectric Field Strength6. Use the Pythagorean theorem to find the

magnitude of the resultant electric field strength vector.

22

, ,

2 25 5

5

1.08 10 N/C 2.49 10 N/C

2.71 10 N/C

tot x tot y tot

tot

tot

E E E

E

E

The Electric Field

Sample Problem, continuedElectric Field Strength7. Use a suitable trigonometric function to find the

direction of the resultant electric field strength vector.In this case, you can use the inverse tangent function:

5,

5,

2.49 10 N/Ctan

1.08 10 N/C

66.0

y tot

x tot

E

E

The Electric Field

Sample Problem, continuedElectric Field Strength8. Evaluate your answer.

The electric field at point P is pointing away from the charge q1, as expected, because q1 is a positive charge and is larger than the negative charge q2.

The Electric Field

The Electric Field

Electric Field Lines

• The number of electric field lines is proportional to the electric field strength.

• Electric field lines are tangent to the electric field vector at any point.

The Electric Field

Conductors in Electrostatic Equilibrium• The electric field is zero everywhere inside the

conductor.

• Any excess charge on an isolated conductor resides entirely on the conductor’s outer surface.

• The electric field just outside a charged conductor is perpendicular to the conductor’s surface.

• On an irregularly shaped conductor, charge tends to accumulate where the radius of curvature of the surface is smallest, that is, at sharp points.

Multiple Choice

1. In which way is the electric force similar to the gravitational force?

A. Electric force is proportional to the mass of the object.

B. Electric force is similar in strength to gravitational force.

C. Electric force is both attractive and repulsive.

D. Electric force decreases in strength as the distance between the charges increases.

Multiple Choice, continued

1. In which way is the electric force similar to the gravitational force?

A. Electric force is proportional to the mass of the object.

B. Electric force is similar in strength to gravitational force.

C. Electric force is both attractive and repulsive.

D. Electric force decreases in strength as the distance between the charges increases.


2. What must the charges be for A and B in the figure so that they produce the electric field lines shown?

F. A and B must both be positive.

G. A and B must both be negative.

H. A must be negative, and B must be positive.

J. A must be positive, and B must be negative.


3. Which activity does not produce the same results as the other three?

A. sliding over a plastic-covered automobile seat

B. walking across a woolen carpet

C. scraping food from a metal bowl with a metal spoon

D. brushing dry hair with a plastic comb


4. By how much does the electric force between two charges change when the distance between them is doubled?

4

2

1 21

4

F.

G.

H.

J.


4. By how much does the electric force between two charges change when the distance between them is doubled?

4

2

1 4

1

2

F.

G.

H.

J.


Use the passage below to answer questions 5–6.

A negatively charged object is brought close to the surface of a conductor, whose opposite side is then grounded.

5. What is this process of charging called?

A. charging by contact

B. charging by induction

C. charging by conduction

D. charging by polarization


Use the passage below to answer questions 5–6.

A negatively charged object is brought close to the surface of a conductor, whose opposite side is then grounded.

6. What kind of charge is left on the conductor’s surface?

F. neutral

G. negative

H. positive

J. both positive and negative


7. What is the electric field strength 2.0 m from the center of the conducting sphere?

A. 0 N/C

B. 5.0 102 N/C

C. 5.0 103 N/C

D. 7.2 103 N/C

Use the graph below to answer questions 7–10. The graph shows the electric field strength at different distances from the center of the charged conducting sphere of a Van de Graaff generator.


8. What is the strength of the electric field at the surface of the conducting sphere?

F. 0 N/C

G. 1.5 102 N/C

H. 2.0 102 N/C

J. 7.2 103 N/C



9. What is the strength of the electric field inside the conducting sphere?

A. 0 N/C

B. 1.5 102 N/C

C. 2.0 102 N/C

D. 7.2 103 N/C



10. What is the radius of the conducting sphere?

F. 0.5 m

G. 1.0 m

H. 1.5 m

J. 2.0 m


Short Response

11. Three identical charges (q = +5.0 mC) are along a circle with a radius of 2.0 m at angles of 30°, 150°, and 270°, as shown in the figure.What is the resultant electric field at the center?

Short Response, continued

11. Three identical charges (q = +5.0 mC) are along a circle with a radius of 2.0 m at angles of 30°, 150°, and 270°, as shown in the figure.What is the resultant electric field at the center?

Answer: 0.0 N/C


12. If a suspended object is attracted to another object that is charged, can you conclude that the suspended object is charged? Briefly explain your answer.


12. If a suspended object is attracted to another object that is charged, can you conclude that the suspended object is charged? Briefly explain your answer.

Answer: not necessarily; The suspended object might have a charge induced on it, but its overall charge could be neutral.


13. One gram of hydrogen contains 6.02 1023 atoms, each with one electron and one proton. Suppose that 1.00 g of hydrogen is separated into protons and electrons, that the protons are placed at Earth’s north pole, and that the electrons are placed at Earth’s south pole. Assuming the radius of Earth to be 6.38 106 m, what is the magnitude of the resulting compressional force on Earth?


13. One gram of hydrogen contains 6.02 1023 atoms, each with one electron and one proton. Suppose that 1.00 g of hydrogen is separated into protons and electrons, that the protons are placed at Earth’s north pole, and that the electrons are placed at Earth’s south pole. Assuming the radius of Earth to be 6.38 106 m, what is the magnitude of the resulting compressional force on Earth?

Answer: 5.12 105 N


14. Air becomes a conductor when the electric field strength exceeds 3.0 106 N/C. Determine the maximum amount of charge that can be carried by a metal sphere 2.0 m in radius.


14. Air becomes a conductor when the electric field strength exceeds 3.0 106 N/C. Determine the maximum amount of charge that can be carried by a metal sphere 2.0 m in radius.

Answer: 1.3 10–3 C

Extended Response

Use the information below to answer questions 15–18.

A proton, which has a mass of 1.673 10–27 kg, accelerates from rest in a uniform electric field of 640 N/C. At some time later, its speed is 1.2 106 m/s.

15. What is the magnitude of the acceleration of the proton?

Extended Response, continued



15. What is the magnitude of the acceleration of the proton?

Answer: 6.1 1010 m/s2




16. How long does it take the proton to reach this speed?




16. How long does it take the proton to reach this speed?

Answer: 2.0 10–5 s




17. How far has it moved in this time interval?




17. How far has it moved in this time interval?

Answer: 12 m




18. What is its kinetic energy at the later time?




18. What is its kinetic energy at the later time?

Answer: 1.2 10–15 J


19. A student standing on a piece of insulating material places her hand on a Van de Graaff generator. She then turns on the generator. Shortly thereafter, her hairs stand on end. Explain how charge is or is not transferred in this situation, why the student is not shocked, and what causes her hairs to stand up after the generator is started.


19. (See previous slide for question.)Answer: The charge on the sphere of the Van de Graaff

generator is transferred to the student by means of conduction. This charge remains on the student because she is insulated from the ground. As there is no path between the student and the generator and the student and the ground by which charge can escape, the student is not shocked. The accumulation of charges of the same sign on the strands of the student’s hair causes the strands to repel each other and so stand on end.

Electric Charge

Charging By Induction

Electric Charge

Transfer of Electric Charge

The Electric Field

Electric Field Lines

Electric Potential

Electrical Potential Energy

• Electrical potential energy is potential energy associated with a charge due to its position in an electric field.

• Electrical potential energy is a component of mechanical energy.

ME = KE + PEgrav + PEelastic + PEelectric

Electric Potential

Electrical Potential Energy, continued

• Electrical potential energy can be associated with a charge in a uniform field.

• Electrical Potential Energy in a Uniform Electric Field

PEelectric = –qEdelectrical potential energy = –(charge) (electric field strength)

(displacement from the reference point in the direction of the field)

Electric Potential

Potential Difference

• Electric Potential equals the work that must be performed against electric forces to move a charge from a reference point to the point in question, divided by the charge.

• The electric potential associated with a charge is the electric energy divided by the charge:

V

PEelectric

q

Electric Potential

Potential Difference, continued

• Potential Difference equals the work that must be performed against electric forces to move a charge between the two points in question, divided by the charge.

• Potential difference is a change in electric potential.

change in electric potential energy

potential differenceelectric charge

electricPEV

q

Electric Potential


• The potential difference in a uniform field varies with the displacement from a reference point.

• Potential Difference in a Uniform Electric Field

∆V = –Ed

potential difference = –(magnitude of the electric field displacement)

Electric Potential

Sample ProblemPotential Energy and Potential Difference

A charge moves a distance of 2.0 cm in the direction of a uniform electric field whose magnitude is 215 N/C.As the charge moves, its electrical potential energy decreases by 6.9 10-

19 J. Find the charge on the moving particle. What is the potential difference between the two locations?

Electric Potential

Sample Problem, continuedPotential Energy and Potential Difference

Given:

∆PEelectric = –6.9 10–19 J

d = 0.020 m

E = 215 N/C

Unknown:

q = ?

∆V = ?

Electric Potential


Use the equation for the change in electrical potential energy.

PEelectric = –qEd

Rearrange to solve for q, and insert values.

–19

–19

(–6.9 10 J)– –

(215 N/C)(0.020 m)

1.6 10 C

electricPEq

Ed

q

Electric Potential


The potential difference is the magnitude of E times the displacement.

– –(215 N/C)(0.020 m)

–4.3 V

V Ed

V

Electric Potential


• At right, the electric poten-tial at point A depends on the charge at point B and the distance r.

• An electric potential exists at some point in an electric field regardless of whether there is a charge at that point.

Electric Potential

Potential Difference, continued• The reference point for potential difference near a

point charge is often at infinity.

• Potential Difference Between a Point at Infinity and a Point Near a Point Charge

• The superposition principle can be used to calculate the electric potential for a group of charges.

value of the point chargepotential difference = Coulomb constant

distance to the point charge

C

qV k

r

Capacitance

Capacitors and Charge Storage

• A capacitor is a device that is used to store electrical potential energy.

• Capacitance is the ability of a conductor to store energy in the form of electrically separated charges.

• The SI units for capacitance is the farad, F, which equals a coulomb per volt (C/V)

Capacitance

Capacitors and Charge Storage, continued

• Capacitance is the ratio of charge to potential difference.

magnitude of charge on each platecapacitance =

potential difference

QC

V

Capacitance


• Capacitance depends on the size and shape of a capacitor.

• Capacitance for a Parallel-Plate Capacitor in a Vacuum

–12 2

0

0

area of one of the platescapacitance = permittivity of a vacuum

distance between the plates

of the medium 8.85 10 C /N mpermittivity

AC

d

Capacitance


• The material between a capacitor’s plates can change its capacitance.

• The effect of a dielectric is to reduce the strength of the electric field in a capacitor.

Capacitors in Keyboards

Capacitance

Capacitance

Energy and Capacitors

• The potential energy stored in a charged capacitor depends on the charge and the potential difference between the capacitor’s two plates.

1electrical potential energy = (charge on one plate)(final potential difference)

2

1

2electricPE Q V

Capacitance

Sample ProblemCapacitance

A capacitor, connected to a 12 V battery, holds 36 µC of charge on each plate. What is the capacitance of the capacitor? How much electrical potential energy is stored in the capacitor?

Given:

Q = 36 µC = 3.6 10–5 C

∆V = 12 V

Unknown:

C = ? PEelectric = ?

Sample Problem, continuedCapacitance

To determine the capacitance, use the definition of capacitance.

–5

–6

3.6 10 C

12 V

3.0 10 F 3.0 µF

QC

V

C

Capacitance

Sample Problem, continuedCapacitance

To determine the potential energy, use the alternative form of the equation for the potential energy of a charged capacitor:

2

–6 2

–4

1( )

21

(3.0 10 F)(12 V)2

2.2 10 J

electric

electric

electric

PE C V

PE

PE

Capacitance

Current and Resistance

Current and Charge Movement

• Electric current is the rate at which electric charges pass through a given area.

charge passing through a given area

electric current = time interval

QI

t


Drift Velocity

• Drift velocity is the the net velocity of a charge carrier moving in an electric field.

• Drift speeds are relatively small because of the many collisions that occur when an electron moves through a conductor.


Resistance to Current

• Resistance is the opposition presented to electric current by a material or device.

• The SI units for resistance is the ohm (Ω) and is equal to one volt per ampere.

• Resistance

potential difference

resistancecurrent

VR

I


Resistance to Current, continued

• For many materials resistance is constant over a range of potential differences. These materials obey Ohm’s Law and are called ohmic materials.

• Ohm’s low does not hold for all materials. Such materials are called non-ohmic.

• Resistance depends on length, cross-sectional area, temperature, and material.


Resistance to Current, continued

• Resistors can be used to control the amount of current in a conductor.

• Salt water and perspiration lower the body's resistance.

• Potentiometers have variable resistance.

Electric Power

Sources and Types of Current

• Batteries and generators supply energy to charge carriers.

• Current can be direct or alternating.– In direct current, charges move in a single

direction.– In alternating current, the direction of charge

movement continually alternates.

Electric Power

Energy Transfer

• Electric power is the rate of conversion of electrical energy.

• Electric power

P = I∆V

Electric power = current potential difference

Electric Power

Energy Transfer, continued

• Power dissipated by a resistor

• Electric companies measure energy consumed in kilowatt-hours.

• Electrical energy is transferred at high potential differences to minimize energy loss.

22 ( )V

P I V I RR

Multiple Choice

1. What changes would take place if the electron moved from point A to point B in the uniform electric field?A. The electron’s electrical potential energy would increase; its electric potential would increase.B. The electron’s electrical potential energy would increase; its electric potential would decrease.C. The electron’s electrical potential energy would decrease; its electric potential would decrease.D. Neither the electron’s electrical potential energy nor its electric potential would change.


1. What changes would take place if the electron moved from point A to point B in the uniform electric field?A. The electron’s electrical potential energy would increase; its electric potential would increase.B. The electron’s electrical potential energy would increase; its electric potential would decrease.C. The electron’s electrical potential energy would decrease; its electric potential would decrease.D. Neither the electron’s electrical potential energy nor its electric potential would change.


2. What changes would take place if the electron moved from point A to point C in the uniform electric field?F. The electron’s electrical potential energy would increase; its electric potential would increase.G. The electron’s electrical potential energy would increase; its electric potential would decrease.H. The electron’s electrical potential energy would decrease; its electric potential would decrease.J. Neither the electron’s electrical potential energy nor its electric potential would change.


Use the following passage to answer questions 3–4.A proton (q = 1.6 10–19 C) moves 2.0 10–6 m in the direction of an electric field that has a magnitude of 2.0 N/C.

3. What is the change in the electrical potential energy associated with the proton?A. –6.4 10–25 JB. –4.0 10–6 VC. +6.4 10–25 JD. +4.0 10–6 V


Use the following passage to answer questions 3–4.A proton (q = 1.6 10–19 C) moves 2.0 10–6 m in the direction of an electric field that has a magnitude of 2.0 N/C.

4. What is the potential difference between the proton’s starting point and ending point?F. –6.4 10–25 JG. –4.0 10–6 VH. +6.4 10–25 JJ. +4.0 10–6 V


5. If the negative terminal of a 12 V battery is grounded, what is the potential of the positive terminal?

A. –12 V

B. +0 V

C. +6 V

D. +12 V


6. If the area of the plates of a parallel-plate capacitor is doubled while the spacing between the plates is halved, how is the capacitance affected?

F. C is doubled

G. C is increased by four times

H. C is decreased by 1/4

J. C does not change


Use the following passage to answer questions 7–8.

A potential difference of 10.0 V exists across the plates of a capacitor when the charge on each plate is 40.0 µC.

7. What is the capacitance of the capacitor?

A. 2.00 10–4 F

B. 4.00 10–4 F

C. 2.00 10–6 F

D. 4.00 10–6 F



A potential difference of 10.0 V exists across the plates of a capacitor when the charge on each plate is 40.0 µC.

8. How much electrical potential energy is stored in the capacitor?

F. 2.00 10–4 J

G. 4.00 10–4 J

H. 2.00 10–6 J

J. 4.00 10–6 J


9. How long does it take 5.0 C of charge to pass through a given cross section of a copper wire if I = 5.0 A?

A. 0.20 s

B. 1.0 s

C. 5.0 s

D. 25 s


10. A potential difference of 12 V produces a current of 0.40 A in a piece of copper wire. What is the resistance of the wire?

F. 4.8 Ω

G. 12 Ω

H. 30 Ω

J. 36 Ω


11. How many joules of energy are dissipated by a 50.0 W light bulb in 2.00 s?

A. 25.0 J

B. 50.0 J

C. 100 J

D. 200 J


12. How much power is needed to operate a radio that draws 7.0 A of current when a potential difference of 115 V is applied across it?

F. 6.1 10–2 W

G. 2.3 100 W

H. 1.6 101 W

J. 8.0 102 W

Short Response

13. Electrons are moving from left to right in a wire. No other charged particles are moving in the wire. In what direction is the conventional current?


13. Electrons are moving from left to right in a wire. No other charged particles are moving in the wire. In what direction is the conventional current?

Answer: right to left


14. What is drift velocity, and how does it compare with the speed at which an electric field travels through a wire?


14. What is drift velocity, and how does it compare with the speed at which an electric field travels through a wire?

Answer: Drift velocity is the net velocity of a charge carrier moving in an electric field. Drift velocities in a wire are typically much smaller than the speeds at which changes in the electric field propagate through the wire.


15. List four factors that can affect the resistance of a wire.


15. List four factors that can affect the resistance of a wire.

Answer: length, cross-sectional area (thickness), temperature, and material

Extended Response

16. A parallel-plate capacitor is made of two circular plates, each of which has a diameter of 2.50 10–3 m. The plates of the capacitor are separated by a space of 1.40 10–4 m.

a. Assuming that the capacitor is operating in a vacuum and that the permittivity of a vacuum (0 = 8.85 10–

12 C2/N•m2) can be used, determine the capacitance of the capacitor.



a. Assuming that the capacitor is operating in a vacuum and that the permittivity of a vacuum (0 = 8.85 10–

12 C2/N•m2) can be used, determine the capacitance of the capacitor.

Answer: 3.10 10–13 F



b. How much charge will be stored on each plate of the capacitor when the capacitor’s plates are connected across a potential difference of 0.12 V?



b. How much charge will be stored on each plate of the capacitor when the capacitor’s plates are connected across a potential difference of 0.12 V?

Answer: 3.7 10–14 C



c. What is the electrical potential energy stored in the capacitor when fully charged by the potential difference of 0.12 V?



c. What is the electrical potential energy stored in the capacitor when fully charged by the potential difference of 0.12 V?

Answer: 2.2 10–15 J



d. What is the potential difference between a point midway between the plates and a point that is 1.10 10–4 m from one of the plates?



d. What is the potential difference between a point midway between the plates and a point that is 1.10 10–4 m from one of the plates?

Answer: 3.4 10–2 V



e. If the potential difference of 0.12 V is removed from the circuit and the circuit is allowed to discharge until the charge on the plates has decreased to 70.7 percent of its fully charged value, what will the potential difference across the capacitor be?



e. If the potential difference of 0.12 V is removed from the circuit and the circuit is allowed to discharge until the charge on the plates has decreased to 70.7 percent of its fully charged value, what will the potential difference across the capacitor be?

Answer: 8.5 10–2 V

Capacitance

Charging a Capacitor

Capacitance

A Capacitor With a Dielectric

Schematic Diagrams and Circuits

Schematic Diagrams

• A schematic diagram is a representation of a circuit that uses lines to represent wires and different symbols to represent components.

• Some symbols used in schematic diagrams are shown at right.


Electric Circuits

• An electric circuit is a set of electrical components connected such that they provide one or more complete paths for the movement of charges.

• A schematic diagram for a circuit is sometimes called a circuit diagram.

• Any element or group of elements in a circuit that dissipates energy is called a load.


Electric Circuits, continued

• A circuit which contains a complete path for electrons to follow is called a closed circuit.

• Without a complete path, there is no charge flow and therefore no current. This situation is called an open circuit.

• A short circuit is a closed circuit that does not contain a load. Short circuits can be hazardous.



• The source of potential difference and electrical energy is the circuits emf.

• Any device that transforms nonelectrical energy into electrical energy, such as a battery or a generator, is a source of emf.

• If the internal resistance of a battery is neglected, the emf equals the potential difference across the source’s two terminals.



• The terminal voltage is the potential difference across a battery’s positive and negative terminals.

• For conventional current, the terminal voltage is less than the emf.

• The potential difference across a load equals the terminal voltage.

Light Bulb


Resistors in Series or in Parallel

Resistors in Series

• A series circuit describes two or more components of a circuit that provide a single path for current.

• Resistors in series carry the same current.

• The equivalent resistance can be used to find the current in a circuit.

• The equivalent resistance in a series circuit is the sum of the circuit’s resistances.

Req = R1 + R2 + R3…


Resistors in Series


Resistors in Series, continued

• Two or more resistors in the actual circuit have the same effect on the current as one equivalent resistor.

• The total current in a series circuit equals the potential difference divided by the equivalent resistance.

eq

VI

R

Sample Problem

Resistors in SeriesA 9.0 V battery is connected to four light bulbs, as shown at right. Find the equivalent resistance for the circuit and the current in the circuit.



Resistors in Series

1. Define

Given:

∆V = 9.0 V

R1 = 2.0 Ω

R2 = 4.0 Ω

R3 = 5.0 Ω

R4 = 7.0 Ω


Unknown:

Req = ?

I = ?

Diagram:


Resistors in Series

2. Plan

Choose an equation or situation: Because the resistors are connected end to end, they are in series. Thus, the equivalent resistance can be calculated with the equation for resistors in series.

Req = R1 + R2 + R3…

The following equation can be used to calculate the current.

∆V = IReq



Resistors in Series

2. Plan, continued

Rearrange the equation to isolate the unknown: No rearrangement is necessary to calculate Req, but ∆V = IReq must be rearranged to calculate the current.

eq

VI

R



Resistors in Series

3. Calculate

Substitute the values into the equation and solve:

9.0 V0.50 A

18.0 Ωeq

VI

R

Substitute the equivalent resistance value into the equation for current.

eq

eq

R = 2.0 Ω + 4.0 Ω + 5.0 Ω + 7.0 Ω

R = 18.0 Ω



Resistors in Series

4. Evaluate

For resistors connected in series, the equivalent resistance should be greater than the largest resistance in the circuit.

18.0 Ω > 7.0 Ω



Resistors in Series, continued

• Series circuits require all elements to conduct electricity

• As seen below, a burned out filament in a string of bulbs has the same effect as an open switch. Because the circuit is no longer complete, there is no current.


Resistors in Parallel

• A parallel arrangement describes two or more com-ponents of a circuit that provide separate conducting paths for current because the components are connected across common points or junctions

• Lights wired in parallel have more than one path for current. Parallel circuits do not require all elements to conduct.


Resistors in Parallel, continued

• Resistors in parallel have the same potential differences across them.

• The sum of currents in parallel resistors equals the total current.

• The equivalent resistance of resistors in parallel can be calculated using a reciprocal relationship

1

Req

1

R1

1

R2

1

R3

...

Sample Problem

Resistors in ParallelA 9.0 V battery is connected to four resistors, as shown at right. Find the equivalent resistance for the circuit and the total current in the circuit.




1. Define

Given:

∆V = 9.0 V

R1 = 2.0 Ω

R2 = 4.0 Ω

R3 = 5.0 Ω

R4 = 7.0 Ω


Unknown:

Req = ?

I = ?

Diagram:


Resistors in Parallel2. Plan

Choose an equation or situation: Because both sides of each resistor are connected to common points, they are in parallel. Thus, the equivalent resistance can be calculated with the equation for resistors in parallel.


The following equation can be used to calculate the current. ∆V = IReq

1

Req

1

R1

1

R2

1

R3

...



2. Plan, continued

Rearrange the equation to isolate the unknown: No rearrangement is necessary to calculate Req; rearrange ∆V = IReq to calculate the total current delivered by the battery.

eq

VI

R




3. Calculate

Substitute the values into the equation and solve:

eq

eq

eq

1 1 1 1 1 = + + +

R 2.0 Ω 4.0 Ω 5.0 Ω 7.0 Ω

1 0.50 0.25 0.20 0.14 1.09 = + + +

R Ω Ω Ω Ω Ω

1 ΩR = = 0.917 Ω

1.09




3. Calculate, continued

Substitute the equivalent resistance value into the equation for current.

9.0 V

0.917 Ω

9.8 A

eq

VI

R

I




4. Evaluate

For resistors connected in parallel, the equivalent resistance should be less than the smallest resistance in the circuit.

0.917 Ω < 2.0 Ω


Complex Resistor Combinations

Resistors Combined Both in Parallel and in Series• Many complex circuits can be understood by isolating

segments that are in series or in parallel and simplifying them to their equivalent resistances.

• Work backward to find the current in and potential difference across a part of a circuit.


Sample Problem

Equivalent ResistanceDetermine the equivalent resistance of the complex circuit shown below.



Equivalent ResistanceReasoning

The best approach is to divide the circuit into groups of series and parallel resistors. This way, the methods presented in Sample Problems A and B can be used to calculate the equivalent resistance for each group.



Equivalent Resistance1. Redraw the circuit as a group of resistors along

one side of the circuit.Because bends in a wire do not affect the circuit, they do not need to be represented in a schematic diagram. Redraw the circuit without the corners, keeping the arrangement of the circuit elements the same.

TIP: For now, disregard the emf source, and work only with the resistances.



Equivalent Resistance

2. Identify components in series, and calcu-late their equivalent resistance.

Resistors in group (a) and(b) are in series.

For group (a): Req = 3.0 Ω + 6.0 Ω = 9.0 Ω

For group (b): Req = 6.0 Ω + 2.0 Ω = 8.0 Ω



Equivalent Resistance3. Identify components in

parallel, and calculate their equivalent resis-tance.Resistors in group (c) are in parallel.

1 1 1 0.12 0.25 0.37

8.0Ω 4.0Ω 1Ω 1Ω 1Ω

2.7 Ω

eq

eq

R

R



Equivalent Resistance4. Repeat steps 2 and 3 until

the resistors in the circuit are reduced to a single equivalent resistance.The remainder of the resistors, group (d), are in series.

9.0Ω 2.7Ω

For group (d

1.0Ω

):

12.7Ω

eq

eq

R

R


Sample Problem

Current in and Potential Difference Across a ResistorDetermine the current in and potential difference across the 2.0 Ω resistor highlighted in the figure below.



Current in and Potential Difference Across a ResistorReasoning

First determine the total circuit current by reducing the resistors to a single equivalent resistance. Then rebuild the circuit in steps, calculating the current and potential difference for the equivalent resistance of each group until the current in and potential difference across the 2.0 Ω resistor are known.



Current in and Potential Difference Across a Resistor1. Determine the equivalent resistance of the circuit.

The equivalent resistance of the circuit is 12.7 Ω, as calculated in the previous Sample Problem.



Current in and Potential Difference Across a Resistor2. Calculate the total current in the circuit.

Substitute the potential difference and equivalent resistance in ∆V = IR, and rearrange the equation to find the current delivered by the battery.

9.0 V0.71 A

12.7 Ωeq

VI

R



3. Determine a path from the equivalent resistance found in step 1 to the 2.0 Ω resistor.

Review the path taken to find the equivalent resistance in the figure at right, and work backward through this path. The equivalent resistance for the entire circuit is the same as the equivalent resistance for group (d). The center resistor in group (d) in turn is the equivalent resistance for group (c). The top resistor in group (c) is the equivalent resistance for group (b), and the right resistor in group (b) is the 2.0 Ω resistor.



Current in and Potential Difference Across a Resistor4. Follow the path determined in step 3, and calculate

the current in and potential difference across each equivalent resistance. Repeat this process until the desired values are found.


Sample Problem, continued4. A. Regroup, evaluate, and calculate.

Replace the circuit’s equivalent resistance with group (d). The resistors in group (d) are in series; therefore, the current in each resistor is the same as the current in the equivalent resistance, which equals 0.71 A. The potential difference across the 2.7 Ω resistor in group (d) can be calculated using ∆V = IR.

Given: I = 0.71 A R = 2.7 ΩUnknown: ∆V = ?

∆V = IR = (0.71 A)(2.7 Ω) = 1.9 V


Sample Problem, continued4. B. Regroup, evaluate, and calculate.

Replace the center resistor with group (c). The resistors in group (c) are in parallel; therefore, the potential difference across each resistor is the same as the potential difference across the 2.7 Ω equivalent resistance, which equals 1.9 V. The current in the 8.0 Ω resistor in group (c) can be calculated using ∆V = IR.

Given: ∆V = 1.9 V R = 8.0 ΩUnknown: I = ?

1.9 V0.24 A

8.0 Ω

VI

R


Sample Problem, continued4. C. Regroup, evaluate, and calculate.

Replace the 8.0 Ω resistor with group (b).The resistors in group (b) are in series; therefore, the current in each resistor is the same as the current in the 8.0 Ω equivalent resistance, which equals 0.24 A.

The potential difference across the 2.0 Ω resistor can be calculated using ∆V = IR.Given: I = 0.24 A R = 2.0 ΩUnknown: ∆V = ?

(0.24 A)(2.0 Ω)

0.48 V

V IR

V

I 0.24 A

Multiple Choice

1. Which of the following is the correct term for a circuit that does not have a closed-loop path for electron flow?

A. closed circuit

B. dead circuit

C. open circuit

D. short circuit


1. Which of the following is the correct term for a circuit that does not have a closed-loop path for electron flow?

A. closed circuit

B. dead circuit

C. open circuit

D. short circuit


2. Which of the following is the correct term for a circuit in which the load has been unintentionally bypassed?

F. closed circuit

G. dead circuit

H. open circuit

J. short circuit


Use the diagram below to answer questions 3–5.

3. Which of the circuit elements contribute to the load of the circuit?

A. Only A

B. A and B, but not C

C. Only C

D. A, B, and C



4. Which of the following is the correct equation for the equivalent resis-tance of the circuit?

.

1 1 1.

.

1 1 1 1.

eq A B

eq A B

eq

eq A B C

R R R

R R R

R I V

R R R R

F

G

H

J




1 1 1.

.

.

1 1 1 1.

e

eq A B

eq

q A B

eq A B C

R R R

R I V

R R

R R R

R R

G

F

H

J



5. Which of the following is the correct equation for the current in the resistor?

A B C

Beq

B total A

BB

I I I I

VI

R

I I I

VI

R

A.

B.

C.

D.



5. Which of the following is the correct equation for the current in the resistor?

A B C

B

Beq

total A

BB

VI

R

I I I I

I I I

VI

R

A.

C.

D.

B.




–1

.

1 1 1 1.

.

1 1.

eq A B C

eq A B C

eq

eq AB C

R R R R

R R R R

R I V

R RR R

F

G

H

J




–1

1 1.

.

1 1 1 1.

.

eq A B C

eq A B C

B C

eq

eq AR RR

R R R R

R R R R

R I V

R

F

G

H

J



7. Which of the following is the correct equation for the current in resistor B?

A B C

Beq

B total A

BB

B

I I I I

VI

R

I I I

VI

R

A.

B.

C.

D.



7. Which of the following is the correct equation for the current in resistor B?

A B C

Beq

B total

BB

B

A

I I I I

VI

R

VI

R

I I I

A.

B.

C.

D.


8. Three 2.0 Ω resistors are connected in series to a 12 V battery. What is the potential difference across each resistor?

F. 2.0 V

G. 4.0 V

H. 12 V

J. 36 V



Six light bulbs are connected in parallel to a 9.0 V battery. Each bulb has a resistance of 3.0 Ω.

9. What is the potential difference across each bulb?

A. 1.5 V

B. 3.0 V

C. 9.0 V

D. 27 V




10. What is the current in each bulb?

F. 0.5 A

G. 3.0 A

H. 4.5 A

J. 18 A




11. What is the total current in the circuit?

A. 0.5 A

B. 3.0 A

C. 4.5 A

D. 18 A

Short Response

12. Which is greater, a battery’s terminal voltage or the same battery’s emf? Explain why these two quantities are not equal.


12. Which is greater, a battery’s terminal voltage or the same battery’s emf? Explain why these two quantities are not equal.

Answer: A battery’s emf is slightly greater than its terminal voltage. The difference is due to the battery’s internal resistance.


13. Describe how a short circuit could lead to a fire.


13. Describe how a short circuit could lead to a fire.

Answer: In a short circuit, the equivalent resistance of the circuit drops very low, causing the current to be very high. The higher current can cause wires still in the circuit to overheat, which may in turn cause a fire in materials contacting the wires.


14. Explain the advantage of wiring the bulbs in a string of decorative lights in parallel rather than in series.


14. Explain the advantage of wiring the bulbs in a string of decorative lights in parallel rather than in series.

Answer: If one bulb is removed, the other bulbs will still carry current.

Extended Response

15. Using standard symbols for circuit elements, draw a diagram of a circuit that contains a battery, an open switch, and a light bulb in parallel with a resistor. Add an arrow to indicate the direction of current if the switch were closed.


15. Using standard symbols for circuit elements, draw a diagram of a circuit that contains a battery, an open switch, and a light bulb in parallel with a resistor. Add an arrow to indicate the direction of current if the switch were closed.

Answer:



16. For the circuit shown, calculate the following:

a. the equivalent resistance of the circuit

b. the current in the light bulb.

Show all your work for both calculations.



16. For the circuit shown, calculate the following:

a. the equivalent resistance of the circuit

b. the current in the light bulb.

Show all your work for both calculations.

Answer: a. 4.2 Ω b. 2.9 A



17. After a period of time, the 6.0 Ω resistor fails and breaks. Describe what happens to the brightness of the bulb. Support your answer.



17. Answer: The bulb will grow dim. The loss of the 6.0 Ω resistor causes the equivalent resistance of the circuit to increase to 4.5 Ω. As a result, the current in the bulb drops to 2.7 A, and the brightness of the bulb decreases.


18. Find the current in and potential difference across each of the resistors in the following circuits:

a. a 4.0 Ω and a 12.0 Ω resistor wired in series with a 4.0 V source.

b. a 4.0 Ω and a 12.0 Ω resistor wired in parallel with a 4.0 V source.

Show all your work for each calculation.



a. a 4.0 Ω and a 12.0 Ω resistor wired in series with a 4.0 V source.

b. a 4.0 Ω and a 12.0 Ω resistor wired in parallel with a 4.0 V source.

Show all your work for each calculation.Answers: a. 4.0 Ω: 0.25 A, 1.0 V

12.0 Ω: 0.25 A, 3.0 Vb. 4.0 Ω: 1.0 A, 4.0 V

12.0 Ω: 0.33 A, 4.0 V



a. a 150 Ω and a 180 Ω resistor wired in series with a 12 V source.

b. a 150 Ω and a 180 Ω resistor wired in parallel with a 12 V source.



Answer: a.150 Ω: 0.036 A, 5.4 V 180 Ω: 0.036 A, 6.5 Vb. 150 Ω: 0.080 A, 12 V 180 Ω: 0.067 A, 12 V


a. a 150 Ω and a 180 Ω resistor wired in series with a 12 V source.

b. a 150 Ω and a 180 Ω resistor wired in parallel with a 12 V source.


Diagram Symbols


electric charge properties of electric charge there are two kinds of electric charge. –like...

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